Question

The foci of an ellipse are $S(-1,-1),S^{\prime }(0,-2)$ and its $e=\frac{1}{2}$, then the equation of the directrix corresponding to the focus $S$ is :

• A. $x-y+3=0$
• B. $x-y+7=0$
• C. $x-y+5=0$
• D. $x-y+4=0$

Answer 1

The correct option is A $x-y+3=0$

Given, foci of an ellipse are $S(-1,-1)$ and $S^{{}^{\prime }}(0,-2)$ and eccentricity $e=\frac{1}{2}$
We know that distance between foci is $2ae=\sqrt{{(0+1)}^2+{(-2+1)}^2}=\sqrt{2}\Rightarrow a=\sqrt{2}$
Directrix and line joining foci are perpendicular to each other,

$\therefore$ Slope of directrix is $(-\frac{-2+1}{0+1})=1$ and it will be parallel to line at point $S$ with this slope.
Equation of line at $S$ is $\frac{y+1}{x+1}=1⟹x-y=0$, then equation of directrix will be $x-y+k=0$ and its distance from point $S$ is $a(\frac{1}{e}-e)=\frac{3}{\sqrt{2}}=\frac{-1-(-1)+k}{\sqrt{2}}⟹k=3$
$\therefore$ Equation of directrix of ellipse is $x-y+3=0$